Any idea how to find at least one "n" such that:
[ ( x1/n + y1/n ) / 2 ]2n - xy < epsilon
2 <= x <= a, 2 <= y <= a
for any epsilon?
Actually, its well-known that:
lim [ ( x1/n + y1/n ) / 2 ]2n = xy
Follow Math Help Forum on Facebook and Google+
Even for lim n^(1/n) = 1 it is a problem ...
(1/n) (log n) < log(1 + eps) , n > ?? ...
Last edited by yms; November 14th 2012 at 05:30 PM.
View Tag Cloud